18 个 Python 高效编程小技巧,Mark!
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· 2021-12-25
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>>>a=3
>>>b=6
>>>a,b=b,a
>>>print(a)>>>6
>>>ptint(b)>>>5
02 字典推导(Dictionary comprehensions)和集合推导(Set comprehensions)
>>> some_list = [1, 2, 3, 4, 5]
>>> another_list = [ x + 1 for x in some_list ]
>>> another_list
[2, 3, 4, 5, 6]
>>> # Set Comprehensions
>>> some_list = [1, 2, 3, 4, 5, 2, 5, 1, 4, 8]
>>> even_set = { x for x in some_list if x % 2 == 0 }
>>> even_set
set([8, 2, 4])
>>> # Dict Comprehensions
>>> d = { x: x % 2 == 0 for x in range(1, 11) }
>>> d
{1: False, 2: True, 3: False, 4: True, 5: False, 6: True, 7: False, 8: True, 9: False, 10: True}
>>> my_set = {1, 2, 1, 2, 3, 4}
>>> my_set
set([1, 2, 3, 4])
03 计数时使用Counter计数对象
>>> from collections import Counter
>>> c = Counter( hello world )
>>> c
Counter({ l : 3, o : 2, : 1, e : 1, d : 1, h : 1, r : 1, w : 1})
>>> c.most_common(2)
[( l , 3), ( o , 2)]
04 漂亮的打印出JSON
>>> import json
>>> print(json.dumps(data)) # No indention
{"status": "OK", "count": 2, "results": [{"age": 27, "name": "Oz", "lactose_intolerant": true}, {"age": 29, "name": "Joe", "lactose_intolerant": false}]}
>>> print(json.dumps(data, indent=2)) # With indention
{
"status": "OK",
"count": 2,
"results": [
{
"age": 27,
"name": "Oz",
"lactose_intolerant": true
},
{
"age": 29,
"name": "Joe",
"lactose_intolerant": false
}
]
}
写一个程序,打印数字1到100,3的倍数打印“Fizz”来替换这个数,5的倍数打印“Buzz”,对于既是3的倍数又是5的倍数的数字打印“FizzBuzz”。
for x in range(1,101):
print"fizz"[x%3*len( fizz )::]+"buzz"[x%5*len( buzz )::] or x
print "Hello" if True else "World"
>>> Hello
nfc = ["Packers", "49ers"]
afc = ["Ravens", "Patriots"]
print nfc + afc
>>> [ Packers , 49ers , Ravens , Patriots ]
print str(1) + " world"
>>> 1 world
print `1` + " world"
>>> 1 world
print 1, "world"
>>> 1 world
print nfc, 1
>>> [ Packers , 49ers ] 1
x = 2
if 3 > x > 1:
print x
>>> 2
if 1 < x > 0:
print x
>>> 2
nfc = ["Packers", "49ers"]
afc = ["Ravens", "Patriots"]
for teama, teamb in zip(nfc, afc):
print teama + " vs. " + teamb
>>> Packers vs. Ravens
>>> 49ers vs. Patriots
teams = ["Packers", "49ers", "Ravens", "Patriots"]
for index, team in enumerate(teams):
print index, team
>>> 0 Packers
>>> 1 49ers
>>> 2 Ravens
>>> 3 Patriots
numbers = [1,2,3,4,5,6]
even = []
for number in numbers:
if number%2 == 0:
even.append(number)
numbers = [1,2,3,4,5,6]
even = [number for number in numbers if number%2 == 0]
teams = ["Packers", "49ers", "Ravens", "Patriots"]
print {key: value for value, key in enumerate(teams)}
>>> { 49ers : 1, Ravens : 2, Patriots : 3, Packers : 0}
items = [0]*3
print items
>>> [0,0,0]
teams = ["Packers", "49ers", "Ravens", "Patriots"]
print ", ".join(teams)
>>> Packers, 49ers, Ravens, Patriots
data = { user : 1, name : Max , three : 4}
try:
is_admin = data[ admin ]
except KeyError:
is_admin = False
data = { user : 1, name : Max , three : 4}
is_admin = data.get( admin , False)
x = [1,2,3,4,5,6]
#前3个
print x[:3]
>>> [1,2,3]
#中间4个
print x[1:5]
>>> [2,3,4,5]
#最后3个
print x[3:]
>>> [4,5,6]
#奇数项
print x[::2]
>>> [1,3,5]
#偶数项
print x[1::2]
>>> [2,4,6]
from collections import Counter
print Counter("hello")
>>> Counter({ l : 2, h : 1, e : 1, o : 1})
from itertools import combinations
teams = ["Packers", "49ers", "Ravens", "Patriots"]
for game in combinations(teams, 2):
print game
>>> ( Packers , 49ers )
>>> ( Packers , Ravens )
>>> ( Packers , Patriots )
>>> ( 49ers , Ravens )
>>> ( 49ers , Patriots )
>>> ( Ravens , Patriots )
False = True
if False:
print "Hello"
else:
print "World"
>>> Hello
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